###### angle

What is the largest angle in the triangle with sides 5, 12 and 13?

First of all we should check if the triangle is rectangular.

$$5^2+12^2=13^2$$

$$25+144=169$$

169=169

The triangle is rectangular, then the biggest angle will be 90o

###### trapezium

Bases of the trapezium are 7 cm and 15 cm, and a height is 8 cm. Find the area of a trapezium.

$$S=\frac{7+15}{2}\times8=22\times4=88 sm^2$$

###### Triangle

The hypotenuse AB of a right triangle ABC is 8 cm, the angle A is 60o. Find the legs and an acute angle.

"a" is the leg opposite the 60° angle. "b" is the other leg.

sin60=$$\frac{\sqrt{3}}{2}=\frac{a}{8}$$; $$a=4\sqrt{3}$$ cm

cos60=0,5=b\8; b=4 cm.

180°-(90°+60°)=30° is the last angle.

###### square

The area of a square with a diagonal of 6 cm, is

"a" is the side of the square. Two sides and the diagonal is a right triangle. Then $$a^2+a^2=6^2; 2a^2=36; a^2=18$$

The area of the square is $$S=a^2=18 \ cm^2$$

The length of the rectangle is increased by 25%. What percent is necessary to reduce the width to its area has not changed?

"a" is the length, "b" is the width.

S=ab at the begining.

The new area is S=1,25a(xb)=ab.

$$x=\frac{1}{1,25}=0,8$$

The width was reduced by 20%.

###### trigonometry

$$sin\frac{3\pi}{4}\times{cos\frac{\pi}{4}}+cos\frac{3\pi}{4}\times{sin\frac{\pi}{4}}$$

Compute

$$sin\frac{3\pi}{4}\times{cos\frac{\pi}{4}}+cos\frac{3\pi}{4}\times{sin\frac{\pi}{4}}==sin(\frac{3\pi}{4}+\frac{\pi}{4})=sin\pi=0$$

The area of the triangle is 40 cm2. Height is 5 times lower then the side on which it is lowered. Then the height is

Area: S=0,5ah=40

where "a" is for the side and "h" is the height. (h>0; a>0)

Then a=5h.

S=0,5h(5h)=40

$$5h^2=80$$

$$h^2=16$$; h=4

###### rhombus

Find the perimeter of a rhombus, if its side is 2 m

Rhombus has equal sides, then its perimeter is P=4*2=8 m

###### Vector 1

The vector $$\overrightarrow{c}$$(2m;-3) is the sum of the vectors $$\overrightarrow{a}$$(7m;5) and $$\overrightarrow{b}$$(-18;2n) when

Find m and n

2m=7m-18 => m=3,6

-3=5+2n => n=-4

###### trapezium 2

The middle line of the trapezoid is equal to 7 cm. One of the bases is more than another for 4 cm. Find the bases of the trapezoid.

"a" and "b" are the bases of the trapezium.

b=a+4

(a+a+4)/2=7

a+2=7

a=5

b=9

###### expression 23

$$ctg\beta-\frac{cos\beta-1}{sin\beta}$$

Simplify

$$ctg\beta-\frac{cos\beta-1}{sin\beta}=\frac{cos\beta}{sin\beta}-\frac{cos\beta-1}{sin\beta}=\frac{cos\beta-cos\beta+1}{sin\beta}=\frac{1}{sin\beta}$$

###### Parallelogram

One of the angles of the parallelogram constitutes a quarter of the other, then the smaller angle is equal to

"a" and "b" are the angles of the parallelogram.

a=0,25b; b=4a

a+b=180

a+4a=180; 5a=180; a=36°

###### Triangle 4

If the sum of the two acute angles of a triangle is equal to 90, then a triangle is

$$\angle{A}+\angle{B}=90$$°

$$\angle{A}+\angle{B}+\angle{C}=180$$°

$$\angle{C}=180-90=90$$°

The triangle is rectangular.

###### angle 2

Obtuse angle of the parallelogram is 130°. Find the outside angle at an acute angle of the parallelogram.

The outer angle will be equal to the obtuse angle, 130°.

###### area

Find the area of the triangle if its height is 10 cm, its base is 5 cm.

$$S=\frac{1}{2}ah=0,5\times10\times5=25 cm^2$$

###### rectangle

One of the sides of the rectangle is 5 cm larger than the other. Find the sides of the rectangle, if its area is 14 cm2.

a - one side.

a+5 - another side.

a(a+5)=14

$$a^2+5a-14=0$$

a=-7 (not fit) or

a=2; a+5=7

###### square 2

The area of a square is equal to 8 cm. Then its diagonal is

a - the side of the square. d - the diagonal.

a2=8

d2=2a2

$$d=\sqrt{2a^2}=\sqrt{2\times8}=\sqrt{16}=4$$

###### angle 3

For how many degrees 30% of the right angle is less than 20% of the straight angle?

90°*0,3=27°

180°*0,2=36°

36°-27°=9°

###### rectangle 2

The sides of the rectangle are treated as 4 to 9, and its area is 144 cm2. Then a large side is

a and b - the sides.

a/b=4/9; a=4b/9

ab=144

4b2/9=144

b2=324

b=18; a=8

###### polygon

How many sides does a regular polygon have if its every angle is 150°?

$$\frac{180(n-2)}{n}=150$$

180n-360=150n

30n=360

n=12

###### angle 4

One of the angles of a parallelogram is equal to 40°. Find the angle that not equal to it.

The sum of twa unequal angles is 180°.

180°-40°=140°

###### rhombus 2

The perimeter of the rhombus is equal to 24 cm, and the area is 24 cm2. Then the height of the rhombus is

P=4a=24; a=6

S=ah=24; h=24/a=24/6=4 cm

###### line

The equation of the line passing through the point M (7, -11) and parallel to the line y = -2x + 5 is

y=kx+b - the equation of the line.

Parallel lines have equal k.

Then k=-2

y=-2x+b. It passes through the point M, then:

-11=-2*7+b

b=3

y=-2x+3

###### rhombus 3

The height of the rhombus is equal to 4 cm, and its area is 24 cm2. Then the perimeter of the rhombus is

S=ah=24 cm2

h=4 cm; a=24:h=24:4=6 cm

P=4a=4*6=24 cm

###### rectangle 3

The perimeter of the square is 24 cm. Find the length of the sides of a rectangle having the same perimeter, knowing that one side of the rectangle is 2 times larger than the other.

a and 2a - the sides of the rectangle.

2(a+2a)=24

6a=24

a=4 cm

2a=8 cm

###### segment 2

The segment on the map has a length of 3.2 cm, and on the terrain - 1,6 km. Then the segment corresponding to the distance on the terrain of 2,8 km, on the same map will have a length of

The segment on the map has a length of 3.2 cm, and on the terrain - 1,6 km. Then the segment corresponding to the distance on the terrain of 2,8 km, on the same map will have a length of

x - the distance on the map

$$\frac{3,2}{1,6}=\frac{x}{2,8}$$

$$x = \frac{3,2\times2,8}{1,6}$$

x=5,6

###### Ratio 2

Squares of two similar pentagons are in the ration of 4:9. Find the ratio of their perimeters.

Squares of two similar pentagons are in the ration of 4:9. Find the ratio of their perimeters.

If the squares of the pentagons are in the ratio of a:b, then the perimetres are in the ratio of $$\sqrt{a}:\sqrt{b}$$; or $$\sqrt4:\sqrt9=2:3$$

###### rectangle 4

Find the perimeter of a rectangle, if the perpendiculars drawn from the point of intersection of the diagonals of the sides of a rectangle are equal to 2 and 4 dm.

Find the perimeter of a rectangle, if the perpendiculars drawn from the point of intersection of the diagonals of the sides of a rectangle are equal to 2 dm and 4 dm.

These perpendiculars will be equal to the half of corresponding sides of the rectangle. Then sides of the rectangle will be 4 and 8 dm.

The perimeter is: P=2(4+8)=24 dm

###### rectangle 5

The area of the rectangle is 42 cm2. If the width of the rectangle is 3 cm, then what is its length?

The area of the rectangle is 42 cm2. If the width of the rectangle is 3 cm, then what is its length?

x - is the length of the rectangle.

Then

3x=42;

x=14 cm

###### square 3

The radius of the inscribed circle in a square is 2 cm. Find the area of the square.

The radius of the inscribed circle in a square is 2 cm. Find the area of the square.

The side of the square is equal to 2r, where "r" is the radius of the inscribed circle.

Then

S=a2=(2r)2=42=16 cm2

###### trigonometry 3

$$cos\alpha\times{tg}\alpha-sin\alpha$$

Simplify

$$cos\alpha\times{tg}\alpha-sin\alpha=cos\alpha\times\frac{sin\alpha}{cos\alpha}-sin\alpha=sin\alpha-sin\alpha=0$$

###### rectangle 6

The length of the rectangle is 40 cm and a width is 60% of its length. Find the width of the rectangle.

The length of the rectangle is 40 cm and a width is 60% of its length. Find the width of the rectangle.

40*0,6=24 cm

###### trigonometry 4

Compute cos3000

Compute cos3000

$$cos300^0=cos(360^0-60^0)=cos60^0=\frac12=0,5$$

###### rectangle 7

What is the area of a rectangle, if the diagonal of 6 cm forms with the side an angle of 60.

What is the area of a rectangle, if the diagonal of 6 cm forms with the side an angle of 60.

The sides are "a" and "b"

a=6sin600

b=6cos600

S=ab=$$6sin60^0\times6cos60^0=36\times\frac12\times\frac{\sqrt3}{2}=9\sqrt3$$ cm2

###### trigonometry 5

If cosa=0,8; then tga could be

If cosa=0,8; then tga could be

$$sina=\pm\sqrt{1-cos^2a}=\pm\sqrt{1-\frac{16}{25}}=\pm\frac34$$

$$tga=\frac{sina}{cosa}=\frac{\pm\frac35}{\frac45}=\pm\frac34=\pm0,75$$